我对贝叶斯统计几乎没有经验(尽管我很想深入了解它),但我相信这就是您所追求的:
df1 <- data.frame(temp = seq(100, 600, 0.01),
pred$summary)
ggplotly(
ggplot() +
geom_point(data = GasolineYield,
aes(x = temp, y = yield, fill = batch),
size = 4, shape = 21) +
xlim(100, 600) +
geom_line(data = df1, aes(y = mean, x = temp), col="red") +
geom_ribbon(data = df1, aes(ymin= X2.5., ymax = X97.5., x = temp), alpha = 0.3) +
theme_classic())
来自?pred.zoib的帮助:
summary if TRUE(默认),每个后验的基本总结
预测值,包括平均值、标准差、最小值、最大值、中值、2.5% 和 97.5%
分位数,提供。
这与您正在绘制的内容有些不同,因为实际上是平均值:
rowSums(pred$pred[[1]])/ncol(pred$pred[[1]]
可视化差异:
df <- data.frame(temp = seq(100, 600, 0.01),
yield = (pred$pred[[1]][, 201] + pred$pred[[2]][, 201])/2)
ggplotly(
ggplot() +
geom_point(data = GasolineYield,
aes(x = temp, y = yield, fill = batch),
size = 4, shape = 21) +
xlim(100, 600) +
geom_line(data = df1, aes(y = mean, x = temp), col="red") +
geom_ribbon(data = df1, aes(ymin= X2.5., ymax = X97.5., x = temp), alpha = 0.3) +
geom_line(data = df, aes(y = yield, x = temp), col="blue") +
theme_classic())
一些额外的注意事项:
all.equal(rowSums(pred$pred[[1]])/ncol(pred$pred[[1]]), df1$mean)
#output
TRUE
all.equal(apply(pred$pred[[1]], 1, quantile, probs = 0.025), df1$X2.5.)
#output
TRUE
all.equal(apply(pred$pred[[1]], 1, quantile, probs = 0.975), df1$X97.5.)
#output
TRUE
max、min 等也是如此。
我不确定pred$pred[[2]] 代表什么,但您可以使用上述方法为它生成摘要并像这样绘制它:
df2 <- data.frame(temp = seq(100, 600, 0.01),
mean = apply(pred$pred[[2]], 1, mean),
X97.5. = apply(pred$pred[[2]], 1, quantile, probs = 0.975),
X2.5. = apply(pred$pred[[2]], 1, quantile, probs = 0.025))
让我们同时绘制两者(小心我的 R 在使用 ggplotly 执行此操作时变得无响应):
ggplot() +
geom_point(data = GasolineYield,
aes(x = temp, y = yield, fill = batch),
size = 4, shape = 21) +
xlim(100, 600) +
geom_line(data = df1, aes(y = mean, x = temp), col="red") +
geom_ribbon(data = df1, aes(ymin= X2.5., ymax = X97.5., x = temp), alpha = 0.3) +
geom_line(data = df2, aes(y = mean, x = temp), col="blue") +
geom_ribbon(data = df2, aes(ymin= X2.5., ymax = X97.5., x = temp), alpha = 0.3)+
theme_classic()